Atomic Structure Result Question 82-1
82. Estimate the difference in energy between 1st and 2nd Bohr’s orbit for a hydrogen atom. At what minimum atomic number, a transition from $n=2$ to $n=1$ energy level would result in the emission of X-rays with $l=3.0 \times 10^{-8} \mathrm{~m}$ ? Which hydrogen atom-like species does this atomic number correspond to?
(1993, 5M)
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Answer:
Correct Answer: 82. ( a )
Solution:
- For $H$-atom, the energy of a stationary orbit is determined as
$ \begin{aligned} & E_n=-\dfrac{k}{n^2} \quad \text { where, } k=\text { constant }\left(2.18 \times 10^{-18} \mathrm{~J}\right) \\ \Rightarrow \quad & \Delta E(n=2 \text { to } n=1)=k\left(1-\dfrac{1}{4}\right)=\dfrac{3}{4} k \\ = \quad & 1.635 \times 10^{-18} \mathrm{~J} \end{aligned} $
For a $H$-like species, energy of stationary orbit is determined as
$ E_n=-\dfrac{k Z^2}{n^2} $
where, $Z=$ atomic number
$ \begin{aligned} & \Rightarrow \quad \Delta E=k Z^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right) \\ & \Rightarrow \quad \dfrac{1}{\lambda}=\dfrac{\Delta E}{h c}=\dfrac{k}{h c} Z^2\left(\dfrac{1}{1}-\dfrac{1}{4}\right)=R_{\mathrm{H}} Z^2 \times \dfrac{3}{4} \\ & \Rightarrow \quad Z^2=\dfrac{4}{3 R_{\mathrm{H}} \lambda}=\dfrac{4}{3 \times 1.097 \times 10^7 \times 3 \times 10^{-8}}=4.05 \\ & \Rightarrow \quad Z=2\left(\mathrm{He}^{+}\right) \\ \end{aligned} $
BETA
